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I encountered a problem where I needed to prove that a set ( A ) is closed under multiplication, given the following properties:

  1. $0 \in A$
  2. $1 \in A$
  3. If $x, y \in A$, then $x - y \in A$
  4. If $x \neq 0$, then $\frac{1}{x} \in A$
  5. If $x, y \in A$, then $x +y \in A$

To prove that $xy \in A$ for $x, y \in A$, I followed these steps:

  1. If $x \neq 0$ and $x \neq 1$, since $x - 1 \in A$ and $x \in A$: $ \frac{1}{x-1} \in A \quad \text{and} \quad \frac{1}{x} \in A$

    Therefore:

    $\frac{1}{x-1} - \frac{1}{x} = \frac{x - (x-1)}{x(x-1)} = \frac{1}{x(x-1)} \in A$

    Consequently:

$ x(x-1) \in A \quad \Rightarrow \quad x^2 - x \in A \quad \Rightarrow \quad x^2 \in A $

  1. Similarly for $y$: $ y^2 \in A $

  2. Since $x + y \in A$, we have: $ (x + y)^2 \in A$

    Thus:

$ (x + y)^2 - (x^2 + y^2) = 2xy \in A $

  1. Therefore: $ \frac{1}{2xy} \in A$

    This implies:

$ \frac{1}{2xy}+\frac{1}{2xy} = \frac{1}{xy} \in A \quad \Rightarrow \quad xy \in A $

My Questions:

  1. Why does this algebraic manipulation prove that $xy \in A$?
  2. Is there an alternative logical approach to prove the closure under multiplication for set $A$?

Any additional insights or corrections to my approach would be greatly appreciated.

Thank you!

Oth S
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    Isn't $A$ just the set of rationals? – lulu Jun 30 '24 at 01:05
  • After all, every integer is in $A$, so every reciprocal of every non zero integer is in $A$, and that gets you all the rationals. – lulu Jun 30 '24 at 01:06
  • That's a great observation! The properties given for set $ A $ do indeed align with the properties of the set of rationals $( \mathbb{Q} )$. However, the problem does not explicitly state that $ A $ is the set of rationals. It only provides certain properties that $A $ must satisfy. My goal is to prove that $ A $ is closed under multiplication based on these properties alone. – Oth S Jun 30 '24 at 01:10
  • I sketched a proof that $A$ is the rationals. That proves it is closed under multiplication. – lulu Jun 30 '24 at 01:22
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    @lulu The rationals is the smallest subset of the reals satisfying these properties, but there are many others, and a priori some of the others may not be closed under multiplication. – Alex Kruckman Jun 30 '24 at 02:35
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    I’m not sure why $1$ is a question. You have literally written down a proof that $A$ is closed under multiplication (except you should observe in step $1$ that if $x = 0$ or $x = 1$, the obviously $x^2 \in A$, and in step $4$ you need to consider the case where $xy = 0$, but those are all minor edge cases). Is there a step in the proof that you’re not sure about? – David Gao Jun 30 '24 at 06:06
  • (Technically, I suppose the proof also depends on the base field having characteristic being not $2$. But I assume the question is about a set of reals or complex numbers, so that won’t be an issue.) – David Gao Jun 30 '24 at 06:11
  • @AlexKruckman $A$ is clearly contained in the rationals. ${0,1}$ are rational and none of the operations can produce can irrational out of rationals – lulu Jun 30 '24 at 06:58
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    @lulu The question never says $A$ is the closure of ${0, 1}$ under the operations, just that it is closed under them. By your logic every field would be $\mathbb{Q}$. – David Gao Jun 30 '24 at 07:18
  • @DavidGao How so? I agree that I assumed that $A$ was in $\mathbb R$. The OP should specify if some other setting was involved (we need arithmetic operations after all). If you assume that $A\in \mathbb F$ for some field $F$, then you still get that $A$ is contained in the prime field of $F$. None of the operations can take you out of that prime field and $1+1+\cdots +1$ gets you every element in the prime field. – lulu Jun 30 '24 at 07:30
  • @lulu “None of the operations can take you out of that prime field”. I agree. But you do know the difference between “$A$ is the closure of ${0, 1}$ under the operations” and “$A$ is closed under the operations”, right? You’re only showing the closure of ${0, 1}$ under the operations is $\mathbb{Q}$, which is absolutely true. But $A$ could be strictly larger, for example, $A = \mathbb{Q}(\sqrt{2})$ easily satisfies all the conditions listed. – David Gao Jun 30 '24 at 07:33
  • @DavidGao Ah, agreed. Yes, I assumed that $A$ was the set of values that could be produced out of $0,1$ by repeated use of the rules. I agree that the OP did not assume that, so I will delete my posted "solution". Thanks. – lulu Jun 30 '24 at 07:38
  • See also here and here on rings whose units $\cup,{0}$ comprise a field. $\ \ $ – Bill Dubuque Jun 30 '24 at 08:17

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